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Repost an answer from Zhihu
This problem is not as simple as many people think. There is a buck-boost circuit inside the power bank, so the entire power bank is not a linear circuit at all. The circuit knowledge of high school and junior high school physics cannot solve this problem!
Before answering, I took a quick look at other people's answers and found the following main problems:
(1) Is short-circuiting the input and output terminals of a power bank equivalent to connecting the positive and negative terminals of a lithium battery?
A: No, there will be no short circuit. The output and input of the power bank are not equal to the positive and negative poles of the lithium battery. There are various buck-boost circuits and protection circuits inside the power bank. The input and output ports are only connected to these circuits instead of directly connecting to the lithium battery.
Next, let's analyze this problem theoretically. Suppose I have a 3.7v lithium battery (considered as a constant voltage source, of course, this is not the way to think about it in practice, because the variable voltage phenomenon is ignored):
The 5V boost circuit inside the power bank looks like this (a diode is missing between LC, thanks for the correction)
The circuit for charging the battery cell is roughly like this (only considering the constant voltage process, corresponding to the state when the battery cell is at high power)
If the output and input terminals are short-circuited, the entire circuit looks like this:
The battery in this circuit topology is obviously not short-circuited.
Assuming that the circuit will continue to work once connected (excluding some products that can automatically identify such misoperation and cut off the power), the law of conservation of energy tells us that the entire system will continue to consume power. So where does the power go? Is it dangerous to do this in extreme cases? When the power bank is working normally, the two MOS tubes will continuously switch between on and off, roughly like this:
(Simultaneous) conduction:
(simultaneous) deadline:
Some people may have noticed that when the MOS tube in the boost circuit is turned on, the positive and negative electrodes of the battery cell will be short-circuited through an inductor. According to the reactance properties of the inductor, the circuit is in a state similar to "open circuit" at the beginning. When the maintenance time tends to positive infinity, the battery cell is equivalent to being directly short-circuited. However, during the normal operation of the MOS tube inside the power bank, the switching frequency is generally greater than 100kHz, and many products have reached the MHz level. Each cycle lasts for a very short time, which seems to be relatively safe. However, it cannot be ignored that most of the control ICs used in the power bank currently use PWM modulation for voltage regulation. Therefore, under normal circumstances, the lithium battery cell will not be directly short-circuited, but there is a risk of instantaneous large current discharge.
(2) Since the theoretical circuit will not pose a serious risk, will there be any difference in the actual product?
Below is the application schematic and block diagram of TI's BQ24195 control IC:
The block diagram of BQ24195:
(Source: BQ24195 datasheet, TI reserves copyright)
The circuit configuration above is a bit complicated, and because the MOS is integrated into the control IC, it looks a bit confusing. However, it is not difficult to see that the actual circuit configuration is basically the same as the circuit schematic I drew in the sketch, so the analysis in question (1) is also applicable here.
(3) Why does my power bank display as charging after I connect it to its input and output ports with a USB cable, but the power level is decreasing, and some devices even become very hot?
In general, if the input port and the output port are connected with a USB cable, the control IC will drive the MOS tube to work all the time in order to maintain the 5v output port and the 4.2v battery port charging voltage. The ripple generated during the operation will return to the power ground through the capacitor and eventually return to the negative electrode of the battery. In other words, under ideal conditions, the power bank will enter a state of slow discharge (ripple can be understood as a small voltage fluctuation, which is only a very small part of the output voltage). The status light shows that it is in charging state because there is voltage input at the input end, but in fact the power of the power bank will not increase.
The above analysis is ideal. The actual circuit is not configured like the BQ24195 DEMO. The quality of the components used may not be the same. Therefore, the discharge speed of the battery may be much faster than the ideal situation, resulting in heating. However, a normal power bank will have certain current limiting measures on the output circuit, and the lithium battery will also have an independent protection circuit, so the lithium battery will not explode.
If we discuss this issue in depth at the technical level, current power banks generally use synchronous buck (asynchronous in the schematic diagram) to improve efficiency. If the duty cycle modification of the switch tube is simply inverted without considering this prank, the buck will enter DCM from CCM when the duty cycle of the upper tube is very small. In addition, the lithium battery cell is equivalent to a boost topology. At this time, the whole circuit will oscillate, and the inductive components will discharge. The capacitor in the buck is very easy to be broken down. If the power bank uses electrolytic capacitors, they may explode, and then it will be a joke.
To sum up, the problem of connecting the input and output ports of a power bank with a wire is not a "short circuit", but if the power bank cannot automatically identify and automatically disconnect the output, or even gets seriously hot, please disconnect it immediately.
(PS: Before answering, I have conducted experiments on many different models of power banks from different manufacturers, and all of them worked fine)
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